Just yesterday I read about this technique for the first time and I'd be lying if I said I fully understood it. I do realize that it involves separate chains on the same digit.

I do puzzles with pencil and paper only. Is this technique explainable just based on relationships among the numbers, without getting into colors? It certainly seems like the principle of simple coloring won't work, i.e., eliminate a candidate that is a buddy of both a "+" and a "-", since with two chains, different results can be obtained depending on whether you start off the chain with one sign or the other. For example, the two digits in c9 could be both "+", both "-" or one of each.

I am currently working on a puzzle that looks like there might be potential for multi-coloring. There are two very short, three-link chains on the same digit. One chain is shown in lower case, the other in upper. Are there any cells where candidates could be eliminated based on these two small chains?

That is, you color each strong link alternating with capital and small letters, continuing as far as possible (like simple coloring with + and -), and use a new letter, for the next group of connected strong links.
Now you can see, that a and b share a unit, That means that not both of them can be true, i.e either A or B must be true. This allows you to eliminate all candidates, that see both A and B, but in this sample there cannot be a candidate in r1c4 or r6c6, because Bb and Aa would not be strong links then.

This is an old sample, where Tracy has made an elimination with multiple coloring:

Simple coloring only uses the fact that (A or a), (B or b) and (c or C) must be true, i.e all candidates that see both (A and a) or (B and b) or (C and c) can be eliminated - none here.

Multiple coloring also uses the connections between the groups:
Since Ab, Ac and bC each share a unit (and cannot both be true), one of (a or B), (a or C) and (B or c) must always be true, i.e. all candidates that see both (a and B) or (a and C) or (B and c) also can be eliminated.
The only candidate in this case is r6c6, which sees both B and c.